Physics · Unit 1 · 5 hrs
Rotational Dynamics and Oscillatory Motion
Exam-focused notes for Rotational Dynamics and Oscillatory Motion (Physics, PHY118): what the TU syllabus asks and how it has actually been tested, with 12 solved past questions from this unit.
What this unit covers
- Moment of inertia and torque
- Rotational kinetic energy
- Conservation of angular momentum
- Oscillation of spring: frequency, period, amplitude, phase angle and energy
Oscillation of spring
An oscillating block of mass 250 g takes 0.15 sec to move between the endpoints of the motion, which are 40 cm apart. (a) What is the frequency of the motion? (b) What is the amplitude of the motion? (c) What is the force constant of the spring? [5]
Quantity Value ------ Mass, $m$ $250 \text{ g} = 0.25 \text{ kg}$ Time between endpoints $0.15 \text{ s}$ Distance between endpoints $40 \text{ cm} = 0.40 \text{ m}$ Key interpretation: Moving between the two endpoints (extreme to extreme) is half of a full...
Full solved answer →An oscillating block of mass 250 g takes 0.2 sec to move between the endpoints of the motion, which are 50 cm apart. Find the frequency and amplitude of the motion. What is the force constant of the spring? [5]
Quantity Value ------ Mass of block $m = 250\text{ g} = 0.25\text{ kg}$ Time to move between endpoints $t = 0.2\text{ s}$ Distance between endpoints $d = 50\text{ cm} = 0.50\text{ m}$ --- Moving from one endpoint to the other is half a full oscillation: $$t...
Full solved answer →Set up differential equation for an oscillation of a spring using Hooke's and Newton's second law. [5]
Consider a spring of negligible mass with one end fixed. A particle of mass m is attached to the free end. Let l be the extension (displacement) of the spring from its natural (equilibrium) position. --- When the spring is stretched or compressed by a displ...
Full solved answer →An oscillating block of mass 250 g takes 0.15 sec to move between the endpoints of the motion, which are 40 cm apart. Find (a) frequency and (b) amplitude of the motion, and (c) force constant of the spring. [5]
Quantity Value ------ Mass $m$ $250\text{ g} = 0.25\text{ kg}$ Time between endpoints $0.15\text{ s}$ Distance between endpoints $40\text{ cm} = 0.40\text{ m}$ --- Moving from one extreme endpoint to the other is half a full oscillation: $$\frac{T}{2} = 0.1...
Full solved answer →A given spring stretches 0.1m when a force of 20N pulls on it. A 2-kg block attached to it on a frictionless surface is pulled to the right 0.2 m and released. (a) What is frequency of oscillation of the block? (b) What are the velocity and acceleration when x=0.12 mx=0.12,mx=0.12m, on the block's first passing this point? [5]
- Spring stretch: $x0 = 0.1$ m under force $F = 20$ N - Mass: $m = 2$ kg - Amplitude: $A = 0.2$ m (pulled right and released from rest) - Frictionless surface - Point of interest: $x = 0.12$ m (first passing) --- $$k = \frac{F}{x0} = \frac{20}{0.1} = 200 \t...
Full solved answer →Set up differential equation for an oscillation of a spring using Hooke's and Newton's second law. Find the general solution of this equation and hence the expressions for period, velocity and acceleration of oscillation.[10]
--- Consider a spring of negligible mass with one end fixed. A particle of mass m is attached to the free end. When the mass is displaced by a distance x (or l) from its equilibrium position, the spring exerts a restoring force. From Hooke's Law, the restor...
Full solved answer →Set up differential equation for an oscillation of a spring using Hooke's and Newton's second law. Find the general solution of this equation and hence the expressions for period, velocity and acceleration of oscillation.[10]
--- Consider a spring of negligible mass with one end fixed. A particle of mass m is attached to the free end. When the mass is displaced by a distance x (or l) from its equilibrium (natural) position, a restoring force acts on it. From Hooke's Law, the res...
Full solved answer →An oscillating block of mass 250 g takes 0.15 sec to move between the endpoints of the motion, which are 40 cm apart. (a) What is the frequency of the motion? (b) What is the amplitude of the motion? (c) What is the force constant of the spring? [5]
Quantity Value ------ Mass $m$ 250 g = 0.25 kg Time to move between endpoints 0.15 s Distance between endpoints 40 cm = 0.40 m Interpretation: Moving from one extreme endpoint to the other is half a full oscillation, so this time equals $T/2$. $$\frac{T}{2}...
Full solved answer →Rotational kinetic energy
Distinguish rigid and non-rigid body. Derive an expression for rotational kinetic energy and discuss the conditions for conservation of energy. A wheel of radius 0.4 m and moment of inertia $1.2\text{ kg-m}^2$, pivoted at the center, is free to rotate without friction. A rope is wound around it and a 2-kg weight is attached to the rope. When the weight has descended 1.5 m from its starting position, find the rotational velocity of the wheel. [10]
- Radius of wheel: $R = 0.4$ m - Moment of inertia of wheel: $I = 1.2$ kg·m² - Mass of hanging weight: $m = 2$ kg - Distance descended: $h = 1.5$ m - Wheel pivoted at center, frictionless (energy conserved) - Starts from rest - $g = 9.8$ m/s² (standard assu...
Full solved answer →Moment of inertia and torque
A roulette wheel with moment of inertia $I = 0.5\ \mathrm{kgm^2}$ rotating initially at 2 rev/sec coasts to a stop from the constant friction torque of bearing. If the torque is 0.4 Nm, how long does it take to stop? [5]
Quantity Value ------ Moment of inertia, $I$ $0.5\ \text{kg m}^2$ Initial rotation rate, $n0$ $2\ \text{rev/s}$ Final rotation rate, $n$ $0\ \text{rev/s}$ Friction torque, $T$ $0.4\ \text{N m}$ Newton's second law for rotation: $T = I\alpha$, then rotationa...
Full solved answer →A large wheel of radius 0.4 m and moment of inertia 1.2 $\mathrm{kgm^2}$, pivoted at the center, is free to rotate without friction. A rope is wound around it and a 2-kg weight is attached to the rope. When the weight has descended 1.5 m from its starting position (a) what is downward velocity? (b) what is the rotational velocity of the wheel? [5]
Quantity Value ------ Radius of wheel, $r$ 0.4 m Moment of inertia, $I$ 1.2 kg·m² Mass of weight, $m$ 2 kg Distance descended, $h$ 1.5 m $g$ 9.8 m/s² Initial velocity 0 (starts from rest) --- The loss in gravitational PE equals the total gain in kinetic ene...
Full solved answer →Describe moment of inertia and torque for a rotating rigid body. Find the expression for rotational kinetic energy and discuss the conditions for conservation.[10]
--- The inability of a body to change its state of rest or uniform rotational motion by itself is called inertia. For a rotating body, this property is called the Moment of Inertia. For a system of particles with masses m₁, m₂, m₃, ..., mₙ at perpendicular ...
Full solved answer →Make Unit 1 stick
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